Question

The sum of 11 consecutive positive integers is 2002. What is the greatest of these 11 integers?
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Answer 1

a 1 = first number

a 2 = a 1 + 1 = second number

a 3 = a 1 + 2 = third number

a 4 = a 1 + 3 = fourth number

a 5 = a 1 + 4 = fifth number

a 6 = a 1 + 5 = sixth number

a 7 = a 1 + 6 = seventh number

a 8 = a 1 + 7 = eighth numer

a 9 = a 1 + 8 = ninth number

a 10 = a 1 + 9 = tenth number

a 11 = a 1 + 10 = eleventh number



a 1 + a 2 + a 3 + a 4 + a 5 + a 6 + a 7 + a 8 + a 9 + a 10 + a 11 = 2002


a 1 + a 1 + 1 + a 1 + 2 + a 1 + 3 + a 1 + 4 + a 1 + 5 + a 1 + 6 + a 1 + 7 + a 1 + 8 + a 1 + 9 + a 1 + 10 = 2002

11 a 1 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 2002

11 a 1 + 55 = 2002

11 a 1 = 2002 - 55

11 a 1 = 1947 Divide both sides by 11

a 1 = 1947 / 11

a 1 = 177


a2 = a 1 + 1 = 177 + 1 = 178

a 3 = a 1 + 2 = 177 + 2 = 179

a 4 = a 1 + 3 = 177 + 3 = 180

a 5 = a 1 + 4 = 177 + 5 = 181

a 6 = a 1 + 5 = 177 + 5 = 182

a 7 = a 1 + 6 = 177 + 6 = 183

a 8 = a 1 + 7 = 177 + 74 = 184

a 9 = a 1 + 8 = 177 + 8 = 185

a 10 = a 1 + 9 = 177 + 9 = 186

a 11 = a 1 + 10 = 177 + 10 = 187


a 1 + a 2 + a 3 + a 4 + a 5 + a 6 + a 7 + a 8 + a 9 + a 10 + a 11 =

177 + 178 + 179 + 180 + 181 + 182 + 183 + 184 + 185 + 186 + 187 = 2002


The greatest is eleventh number:

a 11 = 187

Answer 2

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Answer 3

Wrong answer